JJC_H2_CHEM_P2 ANS

H2 Chemistry 9746/2 JJ 2009 Preliminary Examination Paper 2 (Suggested Answers) 1. (a) (i) Bent/ angular/ v-shape [1] trigonal planar [1] (ii) P/ kPa V/ m3 PV/ kPa m3 6.0 0.375 2.3 12.0 0.188 2.3 18.0 0.125 2.3 (iii) Since PV is constant, SO2 behaves ideally. [1] (iv) At low pressure, the volume of (SO2) molecules become negligible compared to the volume of gas/container. [1] OR At low pressure , molecules are well spaced-out , resulting in negligible intermolecular attraction. [1] (b) (i) 2SO2 (g) + O2 (g)  2SO3 (g) Initial p.p 10 5 - Eqm p.p 10-2x 5-x 2x At eqm, 10 – 2x + 5 – x + 2x = 11 [1] x = 4 PSO2 = 10 - 2(4) = 2 atm () PO2 = 5 - 4 = 1 atm () PSO3 = 2(4) = 8 atm () (ii) 16) 1 ( 2 8 2 2  pK atm-1 [1] with correct units; ecf from (b)(i) (iii) When pressure is lowered, eqm position is shifted to the left to increase the number of gaseous molecules. [1] Yield/ amount of SO3 decreases. [1] [1] S O O O S O O [1] [1] for 3 correct values to 1 d.p 3(): [2] 1-2(): [1] Page 1 of 6

1. (c) (i) OR (ii) 6E(S=O) = 4E(S=O) + 2(248) – (-197) or 2E(S=O) = 2(248) – (-197) E(S=O) = 346.5 kJ mol -1 or 347 kJ mol-1 [1] Energy/ kJ mol-1 2SO2 (g) + O2 (g) 2SO3 (g) 197  2SO2 (g) + 2O (g) 2(248)  4B.E(S=O) or 4E(S=O)  2S (g) + 6O (g) 6B.E(S=O) or 6E(S=O)  4 : [2] 3-2 : [1] 1-0 : [0] Energy/ kJ mol-1 2SO2 (g) + O2 (g) 3 : [2] 2 : [1] 1-0 : [0] 2SO3 (g) 197  2B.E(S=O) or 2E(S=O)  2(248)  2SO2 (g) + 2O (g) 2. (a) Fe(H2O)6 3+  Fe(H2O)5(OH)2+ + H+ [1] (b) (i) 1s22s22p63s23p63d10 [1] (ii) Colourless [1] because Cu+ does not have unfilled/ partially filled d-orbital [1] Page 2 of 6

2. (c) + CH3C slow H C fast +H C l +F e C l3 FeCl4- C NO2 O NO2 O CH3 NO2 O CH3 [] arrow pointing fr C-H bond to +ve charge [] correct arenium ion [] balanced eqn with FeCl4 - and regeneration of catalyst [] arrow pointing fr the ring to CH3CO + 4 []: 2m 2-3 []: 1m 3. (a) Concentration of vanillin may be followed by measuring the colour intensity of vanillin [1] (b) Using the graph [HCN] = 5.00 mol dm-3, (t1/2)1 = (t1/2)2 = 0.095 min [1] Since half-lives is constant, it is 1st order w.r.t vanillin. [1] (c) When the concentration of HCN increased by 50 times (from 0.1 mol dm-3 to 5.00 mol dm-3), half life decreased by 50 times. [1] Order of reaction with respect to HCN is one. [1] (d) Type of reaction: esterification/ ester formation [1] Structure of R C CH2 O O [1]